Question: Integrate. $ \int -2\sin(x)\,dx $ $=$ $+ C$
We need a function whose derivative is $-2\sin(x)$. We know that the derivative of $\cos(x)$ is $-\sin(x)$, so let's start there: $\dfrac{d}{dx} \cos(x) = -\sin(x)$ Now let's multiply by $2$ : $\dfrac{d}{dx}\left[ 2\cos(x) \right]= -2\sin(x)$ Because finding the integral is the opposite of taking the derivative, this means that: $ \int -2\sin(x)\,dx =2 \cos(x)\, + C$ The answer: $2 \cos(x)\, + C$